Question: You have found the following ages (in years) of all 6 turtles at your local zoo: $ 22,\enspace 21,\enspace 59,\enspace 21,\enspace 73,\enspace 27$ What is the average age of the turtles at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Explanation: Because we have data for all 6 turtles at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $6$ ages and divide by $6$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\mu} = \dfrac{22 + 21 + 59 + 21 + 73 + 27}{{6}} = {37.2\text{ years old}} $ Find the squared deviations from the mean for each turtle. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $22$ years $-15.2$ years $231.04$ years $^2$ $21$ years $-16.2$ years $262.44$ years $^2$ $59$ years $21.8$ years $475.24$ years $^2$ $21$ years $-16.2$ years $262.44$ years $^2$ $73$ years $35.8$ years $1281.64$ years $^2$ $27$ years $-10.2$ years $104.04$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{231.04} + {262.44} + {475.24} + {262.44} + {1281.64} + {104.04}} {{6}} $ $ {\sigma^2} = \dfrac{{2616.84}}{{6}} = {436.14\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{436.14\text{ years}^2}} = {20.9\text{ years}} $ The average turtle at the zoo is 37.2 years old. There is a standard deviation of 20.9 years.